Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define N 1111116

int n,k,f[N];
long long a[N],b[N],l[N],r[N],d[N],fl[N],fr[N];

void modify(int t,int l,int r,int x,int y,int z){
	if (x<=l&&y>=r) {f[t]=z;return;}
	if (x<=(l+r)/2) modify(2*t,l,(l+r)/2,x,y,z);
	if (y>=(l+r)/2+1) modify(2*t+1,(l+r)/2+1,r,x,y,z);
}

int query(int t,int l,int r,int x){
	if (l==r) return f[t];
	if (x<=(l+r)/2) return min(f[t],query(2*t,l,(l+r)/2,x));
	return min(f[t],query(2*t+1,(l+r)/2+1,r,x));
}

int read(){
	int x=0,f=1;char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-')  f=-f;
	for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x*f;
}

int main(){
	n=read(),k=read();
	for (int i=1;i<=n;i++){
		a[i]=a[i-1]+read();b[i]=read();
		if (2*(a[i]-a[i-1])>k&&b[i]==1) {printf("-1\n");return 0;}
		if (b[i]==2) l[i]=0,r[i]=k-1;
		else l[i]=(-2*a[i-1]%k+k)%k,r[i]=(-2*a[i]%k+k)%k;
		d[2*i-1]=l[i];d[2*i]=r[i];
	}
	sort(d+1,d+2*n+1);
	for (int i=1;i<=n;i++)
		l[i]=lower_bound(d+1,d+2*n+1,l[i])-d,
		r[i]=lower_bound(d+1,d+2*n+1,r[i])-d;
	fill(f+1,f+8*n+1,n+1);
	for (int i=n;i>=1;i--){
		int t=query(1,1,2*n,l[i]);
		if (t!=n+1) fl[i]=fl[t]+(d[l[t]]-d[l[i]]+k)%k;
		t=query(1,1,2*n,r[i]);
		if (t!=n+1) fr[i]=fl[t]+(d[l[t]]-d[r[i]]+k)%k;
		if (l[i]<=r[i]) {
			if (1<=l[i]-1) modify(1,1,2*n,1,l[i]-1,i);
			if (r[i]+1<=2*n) modify(1,1,2*n,r[i]+1,2*n,i);
		}
		else if (r[i]+1<=l[i]-1) modify(1,1,2*n,r[i]+1,l[i]-1,i);
	}
	long long ans=1ll<<60;
	for (int i=1;i<=n;i++){
		if (query(1,1,2*n,l[i])>=i) ans=min(ans,fl[i]);
		if (query(1,1,2*n,r[i])>=i) ans=min(ans,fr[i]);
	}
	printf("%lld\n",ans+2*a[n]);
	return 0;
}